Question

In Exercises \(25-28,\) a particle is moving along the \(x\) -axis with position function \(x(t) .\) Find the (a) velocity and (b) acceleration, and (c) describe the motion of the particle for \(t \geq 0\) . $$x(t)=t^{3}-3 t+3$$

Short answer

The velocity of the particle is given by \(v(t)=3t^{2}-3\), its acceleration by \(a(t)=6t\), and the particle is moving to the right when \(t > 1\), stationary at \(t = 1\), moving to left when \(0 \leq t < 1\), and speeding up for all \(t \geq 0\).

Step-by-step solution

Finding the velocity

Velocity is the rate of change of position over time. In mathematical terms, it is the derivative of the position function x(t). So, to find the velocity v(t), differentiate the given position function \(x(t)=t^{3}-3 t+3\). Using the power rule for differentiation, the derivative of \(t^{3}\) is \(3t^{2}\), and the derivative of \(-3t\) is \(-3\). The derivative of the constant \(3\) is \(0\). Adding these up, the velocity function is computed as \(v(t)=3t^{2}-3\).

Finding the acceleration

Acceleration is the rate of change of velocity over time, or in other words, it's the derivative of the velocity function. To find the acceleration function a(t), we differentiate the velocity function \(v(t)=3t^{2}-3\). The derivative of \(3t^{2}\) is \(6t\), and the derivative of \(-3\) is \(0\). So, the acceleration function is computed as \(a(t)=6t\).

Describing motion

Now that we have the velocity and acceleration functions, we can describe the motion of the particle for \(t \geq 0\). The particle's velocity at time t is given by \(v(t)=3t^{2}-3\). Therefore, the particle is moving to the right (its velocity is positive) when \(3t^{2}-3 > 0\), which occurs when \(t > 1\), and is stationary (its velocity is zero) at \(t = 1\), and is moving to left (its velocity is negative) when \(0 \leq t < 1\). The particle's acceleration at time t is given by \(a(t)=6t\), which is always positive for \(t \geq 0\), meaning the particle is always speeding up.

Key concepts

Derivative of Position Function

Understanding how a particle moves along a line requires comprehending the fundamental concept of the derivative of its position function. In calculus, this is the first critical step in analyzing motion.

The position function, denoted as \(x(t)\), represents the location of a particle on, for example, the x-axis, at any given time \(t\). The derivative of this function, \(x'(t)\) or \(v(t)\), provides us with the velocity of the particle. It tells us not only how fast the particle is moving but also in which direction (positive for forward, negative for backward, with respect to the x-axis).

In the given exercise, the position function is \(x(t) = t^3 - 3t + 3\). By applying the power rule of differentiation, we find the velocity function to be \(v(t) = 3t^2 - 3\). This velocity function is crucial because it lets us understand the particle's speed and direction at any moment.

Velocity and Acceleration Calculus

After determining the velocity, the next step in calculus motion problems is finding the acceleration. Acceleration represents the rate at which the velocity changes with respect to time, and it is found by taking the derivative of the velocity function, \(v(t)\).

Acceleration can tell us if the particle is speeding up (positive acceleration), slowing down (negative acceleration), or maintaining a constant speed (zero acceleration). In the exercise, the velocity function \(v(t) = 3t^2 - 3\) is differentiated to obtain the acceleration function \(a(t) = 6t\).

This means the particle is always accelerating along the x-axis for \(t \geq 0\), since \(a(t)\) is always positive in this domain. However, an acceleration value does not directly tell us which direction the particle is moving - it simply tells us how the velocity is changing.

Particle Motion Analysis

Particle motion analysis combines information about velocity and acceleration to provide a full picture of the particle's movement. This includes when and where the particle changes direction, its stops, its speed at any moment, and how the speed is changing.

Using the formulas derived in the exercise, let's analyze the motion of the particle. At \(t = 1\), the velocity equals zero \(v(1) = 0\), indicating a stationary point. For times \(t > 1\), the velocity is positive, implying the particle is moving rightward. For \(0 \leq t < 1\), the velocity is negative, suggesting the particle moves leftward.

With a constant positive acceleration for \(t \geq 0\), the speed of the particle continuously increases, regardless of the direction it moves. Understanding this motion helps in predicting future positions and velocities and is a fundamental aspect of physics and engineering that deals with dynamics and kinematics.